3.746 \(\int \sqrt {\frac {-1+x}{5+3 x}} \, dx\)

Optimal. Leaf size=49 \[ \frac {1}{3} \sqrt {x-1} \sqrt {3 x+5}-\frac {8 \sinh ^{-1}\left (\frac {1}{2} \sqrt {\frac {3}{2}} \sqrt {x-1}\right )}{3 \sqrt {3}} \]

[Out]

-8/9*arcsinh(1/4*6^(1/2)*(-1+x)^(1/2))*3^(1/2)+1/3*(-1+x)^(1/2)*(5+3*x)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1958, 50, 54, 215} \[ \frac {1}{3} \sqrt {x-1} \sqrt {3 x+5}-\frac {8 \sinh ^{-1}\left (\frac {1}{2} \sqrt {\frac {3}{2}} \sqrt {x-1}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(-1 + x)/(5 + 3*x)],x]

[Out]

(Sqrt[-1 + x]*Sqrt[5 + 3*x])/3 - (8*ArcSinh[(Sqrt[3/2]*Sqrt[-1 + x])/2])/(3*Sqrt[3])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 1958

Int[(u_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[(u*(e*(a + b*x
^n))^p)/(c + d*x^n)^p, x] /; FreeQ[{a, b, c, d, e, n, p}, x] && GtQ[b*d*e, 0] && GtQ[c - (a*d)/b, 0]

Rubi steps

\begin {align*} \int \sqrt {\frac {-1+x}{5+3 x}} \, dx &=\int \frac {\sqrt {-1+x}}{\sqrt {5+3 x}} \, dx\\ &=\frac {1}{3} \sqrt {-1+x} \sqrt {5+3 x}-\frac {4}{3} \int \frac {1}{\sqrt {-1+x} \sqrt {5+3 x}} \, dx\\ &=\frac {1}{3} \sqrt {-1+x} \sqrt {5+3 x}-\frac {8}{3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {8+3 x^2}} \, dx,x,\sqrt {-1+x}\right )\\ &=\frac {1}{3} \sqrt {-1+x} \sqrt {5+3 x}-\frac {8 \sinh ^{-1}\left (\frac {1}{2} \sqrt {\frac {3}{2}} \sqrt {-1+x}\right )}{3 \sqrt {3}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 76, normalized size = 1.55 \[ \frac {3 (x-1) \sqrt {3 x+5}-8 \sqrt {3} \sqrt {x-1} \sinh ^{-1}\left (\frac {1}{2} \sqrt {\frac {3}{2}} \sqrt {x-1}\right )}{9 \sqrt {\frac {x-1}{3 x+5}} \sqrt {3 x+5}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(-1 + x)/(5 + 3*x)],x]

[Out]

(3*(-1 + x)*Sqrt[5 + 3*x] - 8*Sqrt[3]*Sqrt[-1 + x]*ArcSinh[(Sqrt[3/2]*Sqrt[-1 + x])/2])/(9*Sqrt[(-1 + x)/(5 +
3*x)]*Sqrt[5 + 3*x])

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fricas [A]  time = 0.41, size = 54, normalized size = 1.10 \[ \frac {1}{3} \, {\left (3 \, x + 5\right )} \sqrt {\frac {x - 1}{3 \, x + 5}} + \frac {4}{9} \, \sqrt {3} \log \left (\sqrt {3} {\left (3 \, x + 5\right )} \sqrt {\frac {x - 1}{3 \, x + 5}} - 3 \, x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1+x)/(5+3*x))^(1/2),x, algorithm="fricas")

[Out]

1/3*(3*x + 5)*sqrt((x - 1)/(3*x + 5)) + 4/9*sqrt(3)*log(sqrt(3)*(3*x + 5)*sqrt((x - 1)/(3*x + 5)) - 3*x - 1)

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giac [B]  time = 0.40, size = 74, normalized size = 1.51 \[ -\frac {8}{9} \, \sqrt {3} \log \relax (2) \mathrm {sgn}\left (3 \, x + 5\right ) + \frac {4}{9} \, \sqrt {3} \log \left ({\left | -\sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2 \, x - 5}\right )} - 1 \right |}\right ) \mathrm {sgn}\left (3 \, x + 5\right ) + \frac {1}{3} \, \sqrt {3 \, x^{2} + 2 \, x - 5} \mathrm {sgn}\left (3 \, x + 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1+x)/(5+3*x))^(1/2),x, algorithm="giac")

[Out]

-8/9*sqrt(3)*log(2)*sgn(3*x + 5) + 4/9*sqrt(3)*log(abs(-sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 2*x - 5)) - 1))*sgn(
3*x + 5) + 1/3*sqrt(3*x^2 + 2*x - 5)*sgn(3*x + 5)

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maple [B]  time = 0.01, size = 76, normalized size = 1.55 \[ -\frac {\sqrt {\frac {x -1}{3 x +5}}\, \left (3 x +5\right ) \left (4 \sqrt {3}\, \ln \left (\sqrt {3}\, x +\frac {\sqrt {3}}{3}+\sqrt {3 x^{2}+2 x -5}\right )-3 \sqrt {3 x^{2}+2 x -5}\right )}{9 \sqrt {\left (3 x +5\right ) \left (x -1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x-1)/(5+3*x))^(1/2),x)

[Out]

-1/9*((x-1)/(5+3*x))^(1/2)*(5+3*x)*(4*ln(3^(1/2)*x+1/3*3^(1/2)+(3*x^2+2*x-5)^(1/2))*3^(1/2)-3*(3*x^2+2*x-5)^(1
/2))/((5+3*x)*(x-1))^(1/2)

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maxima [B]  time = 1.39, size = 80, normalized size = 1.63 \[ \frac {4}{9} \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 3 \, \sqrt {\frac {x - 1}{3 \, x + 5}}}{\sqrt {3} + 3 \, \sqrt {\frac {x - 1}{3 \, x + 5}}}\right ) - \frac {8 \, \sqrt {\frac {x - 1}{3 \, x + 5}}}{3 \, {\left (\frac {3 \, {\left (x - 1\right )}}{3 \, x + 5} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1+x)/(5+3*x))^(1/2),x, algorithm="maxima")

[Out]

4/9*sqrt(3)*log(-(sqrt(3) - 3*sqrt((x - 1)/(3*x + 5)))/(sqrt(3) + 3*sqrt((x - 1)/(3*x + 5)))) - 8/3*sqrt((x -
1)/(3*x + 5))/(3*(x - 1)/(3*x + 5) - 1)

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mupad [B]  time = 3.16, size = 57, normalized size = 1.16 \[ -\frac {8\,\sqrt {3}\,\mathrm {atanh}\left (\sqrt {3}\,\sqrt {\frac {x-1}{3\,x+5}}\right )}{9}-\frac {8\,\sqrt {\frac {x-1}{3\,x+5}}}{3\,\left (\frac {3\,x-3}{3\,x+5}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x - 1)/(3*x + 5))^(1/2),x)

[Out]

- (8*3^(1/2)*atanh(3^(1/2)*((x - 1)/(3*x + 5))^(1/2)))/9 - (8*((x - 1)/(3*x + 5))^(1/2))/(3*((3*x - 3)/(3*x +
5) - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {x - 1}{3 x + 5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1+x)/(5+3*x))**(1/2),x)

[Out]

Integral(sqrt((x - 1)/(3*x + 5)), x)

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